Question: Solve for $x$ : $ 2|x + 1| - 10 = -5|x + 1| + 5 $
Answer: Add $ {5|x + 1|} $ to both sides: $ \begin{eqnarray} 2|x + 1| - 10 &=& -5|x + 1| + 5 \\ \\ { + 5|x + 1|} && { + 5|x + 1|} \\ \\ 7|x + 1| - 10 &=& 5 \end{eqnarray} $ Add ${10}$ to both sides: $ \begin{eqnarray} 7|x + 1| - 10 &=& 5 \\ \\ { + 10} &=& { + 10} \\ \\ 7|x + 1| &=& 15 \end{eqnarray} $ Divide both sides by ${7}$ $ \dfrac{7|x + 1|} {{7}} = \dfrac{15} {{7}} $ Simplify: $ |x + 1| = \dfrac{15}{7}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 1 = -\dfrac{15}{7} $ or $ x + 1 = \dfrac{15}{7} $ Solve for the solution where $x + 1$ is negative: $ x + 1 = -\dfrac{15}{7} $ Subtract ${1}$ from both sides: $ \begin{eqnarray} x + 1 &=& -\dfrac{15}{7} \\ \\ {- 1} && {- 1} \\ \\ x &=& -\dfrac{15}{7} - 1 \end{eqnarray} $ Change the ${ - 1}$ to an equivalent fraction with a denominator of $7$ $ x = - \dfrac{15}{7} {- \dfrac{7}{7}} $ $ x = -\dfrac{22}{7} $ Then calculate the solution where $x + 1$ is positive: $ x + 1 = \dfrac{15}{7} $ Subtract ${1}$ from both sides: $ \begin{eqnarray} x + 1 &=& \dfrac{15}{7} \\ \\ {- 1} && {- 1} \\ \\ x &=& \dfrac{15}{7} - 1 \end{eqnarray} $ Change the ${ - 1}$ to an equivalent fraction with a denominator of $7$ $ x = \dfrac{15}{7} {- \dfrac{7}{7}} $ $ x = \dfrac{8}{7} $ Thus, the correct answer is $x = -\dfrac{22}{7} $ or $x = \dfrac{8}{7} $.